Das*_*aev 13 python opencv image-processing contour edge-detection
我的任务是检测土壤表面的裂缝并计算裂缝的总面积。为此,我使用了Canny边缘检测。
输入图像

结果
下一步是将Canny边缘转换为轮廓,因为我想cv2.mean使用cv2.contourArea功能过滤裂纹并使用函数计算其面积。在这一步上,我遇到了问题。当我使用时:
canny_cracks = cv2.Canny(gray, 100, 200)
contours, _ = cv2.findContours(canny_cracks, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
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由于边缘边缘的孔,它无法正确转换。在这里看到问题

我的问题是如何连接边缘的末端以封闭边缘之间的孔?
注意:我使用了轮廓检测而不应用Canny边缘。问题在于轮廓检测会产生很多噪声,并且不能很好地检测所有裂缝。或者,也许我不知道如何像精巧的边缘那样找到轮廓。
从提供的第二张图片开始,这是我解决此问题的方法:
我们从高斯模糊开始,然后将图像转换为灰度。
image = cv2.imread('5.png')
original = image.copy()
blur = cv2.GaussianBlur(image, (3,3), 0)
gray = cv2.cvtColor(blur, cv2.COLOR_BGR2GRAY)
kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (5,5))
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目的是将土壤边缘与盆边缘隔离。为此,我们使用找出花盆cv2.HoughCircles()的外圆,按比例缩小圆以抓住土壤区域,并使用原始图像的形状创建蒙版。
circle_mask = np.zeros(original.shape, dtype=np.uint8)
circles = cv2.HoughCircles(gray, cv2.HOUGH_GRADIENT, 1.5, 200)
# Convert the (x, y) coordinates and radius of the circles to integers
circles = np.round(circles[0, :]).astype("int")
circle_ratio = 0.85
# Loop over the (x, y) coordinates and radius of the circles
for (x, y, r) in circles:
# Draw the circle, create mask, and obtain soil ROI
cv2.circle(image, (x, y), int(r * circle_ratio), (0, 255, 0), 2)
cv2.circle(circle_mask, (x, y), int(r * circle_ratio), (255, 255, 255), -1)
soil_ROI = cv2.bitwise_and(original, circle_mask)
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我们遍历坐标以找到圆的半径。从这里我们画出最大的外圆。
现在要隔离土壤和盆栽,我们应用比例因子来获得
接下来,我们填充圆圈以获得遮罩,然后将其应用于原始图像以获得土壤ROI。
土壤面膜
土壤投资回报率
你的问题是
如何连接边缘的末端以封闭它们之间的孔?
为此,您可以执行“ morphological transformation使用” cv2.morphologyEx()来封闭孔,从而导致
gray_soil_ROI = cv2.cvtColor(soil_ROI, cv2.COLOR_BGR2GRAY)
close = cv2.morphologyEx(gray_soil_ROI, cv2.MORPH_CLOSE, kernel)
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现在,我们使用最小阈值区域找到轮廓cv2.findContours()并使用cv2.contourArea()进行过滤,以消除诸如岩石之类的小噪声。您可以调整最小面积以控制过滤器强度。
cnts = cv2.findContours(close, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
crack_area = 0
minumum_area = 25
for c in cnts:
area = cv2.contourArea(c)
if area > minumum_area:
cv2.drawContours(original,[c], 0, (36,255,12), 2)
crack_area += area
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最后,求和面积,使我们得出裂缝的总面积
3483.5
import cv2
import numpy as np
image = cv2.imread('5.png')
original = image.copy()
blur = cv2.GaussianBlur(image, (3,3), 0)
gray = cv2.cvtColor(blur, cv2.COLOR_BGR2GRAY)
kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (5,5))
circle_mask = np.zeros(original.shape, dtype=np.uint8)
circles = cv2.HoughCircles(gray, cv2.HOUGH_GRADIENT, 1.5, 200)
# Convert the (x, y) coordinates and radius of the circles to integers
circles = np.round(circles[0, :]).astype("int")
circle_ratio = 0.85
# Loop over the (x, y) coordinates and radius of the circles
for (x, y, r) in circles:
# Draw the circle, create mask, and obtain soil ROI
cv2.circle(image, (x, y), int(r * circle_ratio), (0, 255, 0), 2)
cv2.circle(circle_mask, (x, y), int(r * circle_ratio), (255, 255, 255), -1)
soil_ROI = cv2.bitwise_and(original, circle_mask)
gray_soil_ROI = cv2.cvtColor(soil_ROI, cv2.COLOR_BGR2GRAY)
close = cv2.morphologyEx(gray_soil_ROI, cv2.MORPH_CLOSE, kernel)
cnts = cv2.findContours(close, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
crack_area = 0
minumum_area = 25
for c in cnts:
area = cv2.contourArea(c)
if area > minumum_area:
cv2.drawContours(original,[c], 0, (36,255,12), 2)
crack_area += area
print(crack_area)
cv2.imshow('close', close)
cv2.imshow('circle_mask', circle_mask)
cv2.imshow('soil_ROI', soil_ROI)
cv2.imshow('original', original)
cv2.waitKey(0)
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您可以使用形态关闭。这会缩小白色像素之间的间隙。如果您在下面的脚本中输入您的 Canny 图像,您可以自己尝试一下。
结果:
代码:
import cv2
import numpy as np
# function that handles trackbar changes
def doClose(val):
# create a kernel based on trackbar input
kernel = np.ones((val,val))
# do a morphologic close
res = cv2.morphologyEx(img,cv2.MORPH_CLOSE, kernel)
# display result
cv2.imshow("Result", res)
#load image as grayscale
img = cv2.imread("KbMHp.png",0)
# create window and add trackbar
cv2.namedWindow('Result')
cv2.createTrackbar('KernelSize','Result',0,15,doClose)
# display image
cv2.imshow("Result", img)
cv2.waitKey(0)
cv2.destroyAllWindows()
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